Energy of $n-n$ Quantum state for a hydrogen atom. $=E_{n}=\frac{m_{e} e^{4}}{8 n^{2} E_{o}^{2} h^{2}}$
Where,
$m_{e}=$ mass of electron.
$e=$ charge of an electron.
$n=$ quantum number(Orbit).
$E_{o}=$ absolute permittivity.
$h=$ Planck's Constant.
By evaluating the constants we get,
$E_{n}=\frac{-13.6}{n^{2}} e V$
$\therefore$ Energy of $n=3$ Quantum state,
$E_{3}=\frac{-13.6}{3^{2}} e V$
$E_{3}=-1.51 e V \longrightarrow(1)$
Energy of $n=1$ Quantum state,
$E_{1}=\frac{-13.6}{1^{2}} e V$
$E_{1}=-13.6 e V \longrightarrow(2)$
since this transition happens at the total Express of the kinetic energy of the colliding electron $K$ is:
$K=E_{3}-E_{1}$
$K=-1.51-(-13.6)$
$K=12.1 e V$
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